19.11 problem section 9.3, problem 11

Internal problem ID [2158]
Book : Elementary differential equations with boundary value problems. William F. Trench. Brooks/Cole 2001
Section : Chapter 9 Introduction to Linear Higher Order Equations. Section 9.3. Undetermined Coefficients for Higher Order Equations. Page 495
Problem number : section 9.3, problem 11
Date solved : Sunday, October 06, 2024 at 03:12:02 AM
CAS classification : [[_3rd_order, _linear, _nonhomogeneous]]

Solve

\begin{align*} y^{\prime \prime \prime }-7 y^{\prime \prime }+8 y^{\prime }+16 y&=2 \,{\mathrm e}^{4 x} \left (13+15 x \right ) \end{align*}

19.11.1 Solved as higher order constant coeff ode

Time used: 0.128 (sec)

The characteristic equation is

\[ \lambda ^{3}-7 \lambda ^{2}+8 \lambda +16 = 0 \]

The roots of the above equation are

\begin{align*} \lambda _1 &= -1\\ \lambda _2 &= 4\\ \lambda _3 &= 4 \end{align*}

Therefore the homogeneous solution is

\[ y_h(x)={\mathrm e}^{-x} c_1 +{\mathrm e}^{4 x} c_2 +x \,{\mathrm e}^{4 x} c_3 \]

The fundamental set of solutions for the homogeneous solution are the following

\begin{align*} y_1 &= {\mathrm e}^{-x}\\ y_2 &= {\mathrm e}^{4 x}\\ y_3 &= {\mathrm e}^{4 x} x \end{align*}

This is higher order nonhomogeneous ODE. Let the solution be

\[ y = y_h + y_p \]

Where \(y_h\) is the solution to the homogeneous ODE And \(y_p\) is a particular solution to the nonhomogeneous ODE. \(y_h\) is the solution to

\[ y^{\prime \prime \prime }-7 y^{\prime \prime }+8 y^{\prime }+16 y = 0 \]

Now the particular solution to the given ODE is found

\[ y^{\prime \prime \prime }-7 y^{\prime \prime }+8 y^{\prime }+16 y = -\left (-26-30 x \right ) {\mathrm e}^{4 x} \]

The particular solution is now found using the method of undetermined coefficients.

Looking at the RHS of the ode, which is

\[ -\left (-26-30 x \right ) {\mathrm e}^{4 x} \]

Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is

\[ [\{{\mathrm e}^{4 x} x, {\mathrm e}^{4 x}\}] \]

While the set of the basis functions for the homogeneous solution found earlier is

\[ \{{\mathrm e}^{4 x} x, {\mathrm e}^{-x}, {\mathrm e}^{4 x}\} \]

Since \({\mathrm e}^{4 x}\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes

\[ [\{x^{2} {\mathrm e}^{4 x}, {\mathrm e}^{4 x} x\}] \]

Since \({\mathrm e}^{4 x} x\) is duplicated in the UC_set, then this basis is multiplied by extra \(x\). The UC_set becomes

\[ [\{x^{2} {\mathrm e}^{4 x}, x^{3} {\mathrm e}^{4 x}\}] \]

Since there was duplication between the basis functions in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis function in the above updated UC_set.

\[ y_p = A_{1} x^{2} {\mathrm e}^{4 x}+A_{2} x^{3} {\mathrm e}^{4 x} \]

The unknowns \(\{A_{1}, A_{2}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives

\[ 10 A_{1} {\mathrm e}^{4 x}+6 A_{2} {\mathrm e}^{4 x}+30 A_{2} x \,{\mathrm e}^{4 x} = 2 \,{\mathrm e}^{4 x} \left (13+15 x \right ) \]

Solving for the unknowns by comparing coefficients results in

\[ [A_{1} = 2, A_{2} = 1] \]

Substituting the above back in the above trial solution \(y_p\), gives the particular solution

\[ y_p = 2 x^{2} {\mathrm e}^{4 x}+x^{3} {\mathrm e}^{4 x} \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{-x} c_1 +{\mathrm e}^{4 x} c_2 +x \,{\mathrm e}^{4 x} c_3\right ) + \left (2 x^{2} {\mathrm e}^{4 x}+x^{3} {\mathrm e}^{4 x}\right ) \\ \end{align*}

19.11.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d^{3}}{d x^{3}}y \left (x \right )-7 \frac {d^{2}}{d x^{2}}y \left (x \right )+8 \frac {d}{d x}y \left (x \right )+16 y \left (x \right )=2 \,{\mathrm e}^{4 x} \left (13+15 x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & \frac {d^{3}}{d x^{3}}y \left (x \right ) \\ \square & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{1}\left (x \right ) \\ {} & {} & y_{1}\left (x \right )=y \left (x \right ) \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (x \right ) \\ {} & {} & y_{2}\left (x \right )=\frac {d}{d x}y \left (x \right ) \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (x \right ) \\ {} & {} & y_{3}\left (x \right )=\frac {d^{2}}{d x^{2}}y \left (x \right ) \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} \frac {d}{d x}y_{3}\left (x \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y_{3}\left (x \right )=30 \,{\mathrm e}^{4 x} x +26 \,{\mathrm e}^{4 x}+7 y_{3}\left (x \right )-8 y_{2}\left (x \right )-16 y_{1}\left (x \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (x \right )=\frac {d}{d x}y_{1}\left (x \right ), y_{3}\left (x \right )=\frac {d}{d x}y_{2}\left (x \right ), \frac {d}{d x}y_{3}\left (x \right )=30 \,{\mathrm e}^{4 x} x +26 \,{\mathrm e}^{4 x}+7 y_{3}\left (x \right )-8 y_{2}\left (x \right )-16 y_{1}\left (x \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=\left [\begin {array}{c} y_{1}\left (x \right ) \\ y_{2}\left (x \right ) \\ y_{3}\left (x \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}{\moverset {\rightarrow }{y}}\left (x \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -16 & -8 & 7 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (x \right )+\left [\begin {array}{c} 0 \\ 0 \\ 30 \,{\mathrm e}^{4 x} x +26 \,{\mathrm e}^{4 x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (x \right )=\left [\begin {array}{c} 0 \\ 0 \\ 30 \,{\mathrm e}^{4 x} x +26 \,{\mathrm e}^{4 x} \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -16 & -8 & 7 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & \frac {d}{d x}{\moverset {\rightarrow }{y}}\left (x \right )=A \cdot {\moverset {\rightarrow }{y}}\left (x \right )+{\moverset {\rightarrow }{f}} \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [4, \left [\begin {array}{c} \frac {1}{16} \\ \frac {1}{4} \\ 1 \end {array}\right ]\right ], \left [4, \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-x}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair, with eigenvalue of algebraic multiplicity 2}\hspace {3pt} \\ {} & {} & \left [4, \left [\begin {array}{c} \frac {1}{16} \\ \frac {1}{4} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {First solution from eigenvalue}\hspace {3pt} 4 \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}\left (x \right )={\mathrm e}^{4 x}\cdot \left [\begin {array}{c} \frac {1}{16} \\ \frac {1}{4} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Form of the 2nd homogeneous solution where}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {is to be solved for,}\hspace {3pt} \lambda =4\hspace {3pt}\textrm {is the eigenvalue, and}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is the eigenvector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Note that the}\hspace {3pt} x \hspace {3pt}\textrm {multiplying}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {makes this solution linearly independent to the 1st solution obtained from}\hspace {3pt} \lambda =4 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} {\moverset {\rightarrow }{y}}_{3}\left (x \right )\hspace {3pt}\textrm {into the homogeneous system}\hspace {3pt} \\ {} & {} & \lambda \,{\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}=\left ({\mathrm e}^{\lambda x} A \right )\cdot \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Use the fact that}\hspace {3pt} {\moverset {\rightarrow }{v}}\hspace {3pt}\textrm {is an eigenvector of}\hspace {3pt} A \\ {} & {} & \lambda \,{\mathrm e}^{\lambda x} \left (x {\moverset {\rightarrow }{v}}+{\moverset {\rightarrow }{p}}\right )+{\mathrm e}^{\lambda x} {\moverset {\rightarrow }{v}}={\mathrm e}^{\lambda x} \left (\lambda x {\moverset {\rightarrow }{v}}+A \cdot {\moverset {\rightarrow }{p}}\right ) \\ \bullet & {} & \textrm {Simplify equation}\hspace {3pt} \\ {} & {} & \lambda {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Make use of the identity matrix}\hspace {3pt} \mathrm {I} \\ {} & {} & \left (\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}+{\moverset {\rightarrow }{v}}=A \cdot {\moverset {\rightarrow }{p}} \\ \bullet & {} & \textrm {Condition}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {must meet for}\hspace {3pt} {\moverset {\rightarrow }{y}}_{3}\left (x \right )\hspace {3pt}\textrm {to be a solution to the homogeneous system}\hspace {3pt} \\ {} & {} & \left (A -\lambda \cdot I \right )\cdot {\moverset {\rightarrow }{p}}={\moverset {\rightarrow }{v}} \\ \bullet & {} & \textrm {Choose}\hspace {3pt} {\moverset {\rightarrow }{p}}\hspace {3pt}\textrm {to use in the second solution to the homogeneous system from eigenvalue}\hspace {3pt} 4 \\ {} & {} & \left (\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -16 & -8 & 7 \end {array}\right ]-4\cdot \left [\begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {array}\right ]\right )\cdot {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} \frac {1}{16} \\ \frac {1}{4} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Choice of}\hspace {3pt} {\moverset {\rightarrow }{p}} \\ {} & {} & {\moverset {\rightarrow }{p}}=\left [\begin {array}{c} -\frac {1}{64} \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Second solution from eigenvalue}\hspace {3pt} 4 \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}\left (x \right )={\mathrm e}^{4 x}\cdot \left (x \cdot \left [\begin {array}{c} \frac {1}{16} \\ \frac {1}{4} \\ 1 \end {array}\right ]+\left [\begin {array}{c} -\frac {1}{64} \\ 0 \\ 0 \end {array}\right ]\right ) \\ \bullet & {} & \textrm {General solution of the system of ODEs can be written in terms of the particular solution}\hspace {3pt} {\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=\mathit {C1} {\moverset {\rightarrow }{y}}_{1}+\mathit {C2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+\mathit {C3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+{\moverset {\rightarrow }{y}}_{p}\left (x \right ) \\ \square & {} & \textrm {Fundamental matrix}\hspace {3pt} \\ {} & \circ & \textrm {Let}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {be the matrix whose columns are the independent solutions of the homogeneous system.}\hspace {3pt} \\ {} & {} & \phi \left (x \right )=\left [\begin {array}{ccc} {\mathrm e}^{-x} & \frac {{\mathrm e}^{4 x}}{16} & {\mathrm e}^{4 x} \left (\frac {x}{16}-\frac {1}{64}\right ) \\ -{\mathrm e}^{-x} & \frac {{\mathrm e}^{4 x}}{4} & \frac {{\mathrm e}^{4 x} x}{4} \\ {\mathrm e}^{-x} & {\mathrm e}^{4 x} & {\mathrm e}^{4 x} x \end {array}\right ] \\ {} & \circ & \textrm {The fundamental matrix,}\hspace {3pt} \Phi \left (x \right )\hspace {3pt}\textrm {is a normalized version of}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {satisfying}\hspace {3pt} \Phi \left (0\right )=I \hspace {3pt}\textrm {where}\hspace {3pt} I \hspace {3pt}\textrm {is the identity matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\phi \left (x \right )\cdot \phi \left (0\right )^{-1} \\ {} & \circ & \textrm {Substitute the value of}\hspace {3pt} \phi \left (x \right )\hspace {3pt}\textrm {and}\hspace {3pt} \phi \left (0\right ) \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} {\mathrm e}^{-x} & \frac {{\mathrm e}^{4 x}}{16} & {\mathrm e}^{4 x} \left (\frac {x}{16}-\frac {1}{64}\right ) \\ -{\mathrm e}^{-x} & \frac {{\mathrm e}^{4 x}}{4} & \frac {{\mathrm e}^{4 x} x}{4} \\ {\mathrm e}^{-x} & {\mathrm e}^{4 x} & {\mathrm e}^{4 x} x \end {array}\right ]\cdot \left [\begin {array}{ccc} 1 & \frac {1}{16} & -\frac {1}{64} \\ -1 & \frac {1}{4} & 0 \\ 1 & 1 & 0 \end {array}\right ]^{-1} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )=\left [\begin {array}{ccc} \left (1-4 x \right ) {\mathrm e}^{4 x} & -\frac {4 \,{\mathrm e}^{-x}}{5}+\frac {4 \,{\mathrm e}^{4 x}}{5}-3 \,{\mathrm e}^{4 x} x & \frac {{\mathrm e}^{-x}}{5}-\frac {{\mathrm e}^{4 x}}{5}+{\mathrm e}^{4 x} x \\ -16 \,{\mathrm e}^{4 x} x & \frac {4 \,{\mathrm e}^{-x}}{5}+\frac {{\mathrm e}^{4 x}}{5}-12 \,{\mathrm e}^{4 x} x & -\frac {{\mathrm e}^{-x}}{5}+\frac {{\mathrm e}^{4 x}}{5}+4 \,{\mathrm e}^{4 x} x \\ -64 \,{\mathrm e}^{4 x} x & -\frac {4 \,{\mathrm e}^{-x}}{5}+\frac {4 \,{\mathrm e}^{4 x}}{5}-48 \,{\mathrm e}^{4 x} x & \frac {{\mathrm e}^{-x}}{5}+\frac {4 \,{\mathrm e}^{4 x}}{5}+16 \,{\mathrm e}^{4 x} x \end {array}\right ] \\ \square & {} & \textrm {Find a particular solution of the system of ODEs using variation of parameters}\hspace {3pt} \\ {} & \circ & \textrm {Let the particular solution be the fundamental matrix multiplied by}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right )\hspace {3pt}\textrm {and solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & \circ & \textrm {Take the derivative of the particular solution}\hspace {3pt} \\ {} & {} & \frac {d}{d x}{\moverset {\rightarrow }{y}}_{p}\left (x \right )=\left (\frac {d}{d x}\Phi \left (x \right )\right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot \left (\frac {d}{d x}{\moverset {\rightarrow }{v}}\left (x \right )\right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}\Phi \left (x \right )\right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot \left (\frac {d}{d x}{\moverset {\rightarrow }{v}}\left (x \right )\right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {The fundamental matrix has columns that are solutions to the homogeneous system so its derivative follows that of the homogeneous system}\hspace {3pt} \\ {} & {} & A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+\Phi \left (x \right )\cdot \left (\frac {d}{d x}{\moverset {\rightarrow }{v}}\left (x \right )\right )=A \cdot \Phi \left (x \right )\cdot {\moverset {\rightarrow }{v}}\left (x \right )+{\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Cancel like terms}\hspace {3pt} \\ {} & {} & \Phi \left (x \right )\cdot \left (\frac {d}{d x}{\moverset {\rightarrow }{v}}\left (x \right )\right )={\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & \frac {d}{d x}{\moverset {\rightarrow }{v}}\left (x \right )=\Phi \left (x \right )^{-1}\cdot {\moverset {\rightarrow }{f}}\left (x \right ) \\ {} & \circ & \textrm {Integrate to solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right ) \\ {} & {} & {\moverset {\rightarrow }{v}}\left (x \right )=\int _{0}^{x}\Phi \left (s \right )^{-1}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \\ {} & \circ & \textrm {Plug}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (x \right )\hspace {3pt}\textrm {into the equation for the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\Phi \left (x \right )\cdot \left (\int _{0}^{x}\Phi \left (s \right )^{-1}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \right ) \\ {} & \circ & \textrm {Plug in the fundamental matrix and the forcing function and compute}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (x \right )=\left [\begin {array}{c} \frac {\left (25 x^{3}+50 x^{2}-20 x +4\right ) {\mathrm e}^{4 x}}{5}-\frac {4 \,{\mathrm e}^{-x}}{5} \\ \frac {\left (100 x^{3}+275 x^{2}+20 x -4\right ) {\mathrm e}^{4 x}}{5}+\frac {4 \,{\mathrm e}^{-x}}{5} \\ \frac {2 \left (200 x^{3}+550 x^{2}+55 x +2\right ) {\mathrm e}^{4 x}}{5}-\frac {4 \,{\mathrm e}^{-x}}{5} \end {array}\right ] \\ \bullet & {} & \textrm {Plug particular solution back into general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (x \right )=\mathit {C1} {\moverset {\rightarrow }{y}}_{1}+\mathit {C2} {\moverset {\rightarrow }{y}}_{2}\left (x \right )+\mathit {C3} {\moverset {\rightarrow }{y}}_{3}\left (x \right )+\left [\begin {array}{c} \frac {\left (25 x^{3}+50 x^{2}-20 x +4\right ) {\mathrm e}^{4 x}}{5}-\frac {4 \,{\mathrm e}^{-x}}{5} \\ \frac {\left (100 x^{3}+275 x^{2}+20 x -4\right ) {\mathrm e}^{4 x}}{5}+\frac {4 \,{\mathrm e}^{-x}}{5} \\ \frac {2 \left (200 x^{3}+550 x^{2}+55 x +2\right ) {\mathrm e}^{4 x}}{5}-\frac {4 \,{\mathrm e}^{-x}}{5} \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )=\frac {\left (256+1600 x^{3}+3200 x^{2}+20 \left (-64+\mathit {C3} \right ) x +20 \mathit {C2} -5 \mathit {C3} \right ) {\mathrm e}^{4 x}}{320}+\frac {{\mathrm e}^{-x} \left (5 \mathit {C1} -4\right )}{5} \end {array} \]

19.11.3 Maple trace

`Methods for third order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 3; linear nonhomogeneous with symmetry [0,1] 
trying high order linear exact nonhomogeneous 
trying differential order: 3; missing the dependent variable 
checking if the LODE has constant coefficients 
<- constant coefficients successful`
 

19.11.4 Maple dsolve solution

Solving time : 0.008 (sec)
Leaf size : 29

dsolve(diff(diff(diff(y(x),x),x),x)-7*diff(diff(y(x),x),x)+8*diff(y(x),x)+16*y(x) = 2*exp(4*x)*(13+15*x), 
       y(x),singsol=all)
 

\[ y = \left (x^{3}+x c_3 +2 x^{2}+c_2 \right ) {\mathrm e}^{4 x}+{\mathrm e}^{-x} c_1 \]

19.11.5 Mathematica DSolve solution

Solving time : 0.03 (sec)
Leaf size : 42

DSolve[{D[y[x],{x,3}]-7*D[y[x],{x,2}]+8*D[y[x],x]+16*y[x]==2*Exp[4*x]*(13+15*x),{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to e^{4 x} \left (x^3+2 x^2+\left (-\frac {4}{5}+c_3\right ) x+\frac {4}{25}+c_2\right )+c_1 e^{-x} \]